Ordinary Differential Equations in

Neuroscience with Matlab examples.

• Aim 1- Gain understanding of how to set up

and solve ODE’s

• Aim 2 – Understand how to set up an solve

a simple example of the Hebb rule in 2D

Ear

Tongue

Nose

AAAA

BBBB

Classical Conditioning and Hebb’s rule

“When an axon in cell AAAA is near enough to excite cell BBBB and

repeatedly and persistently takes part in firing it, some

growth process or metabolic change takes place in one or both cells such that AAAA’s efficacy in firing BBBB is increased”

D. O. Hebb (1949)

Our goal at end of class – understand Hebbian

plasticity mathematically

The generalized Hebb rule:

where xi are the inputs

and y the output is assumed linear:

Results in 2D

j

j

j

ii

xwy

yxdt

dw

∑=

=η

x1 x2 x3 x4

w1 w2 w3 w4

y

-2 -1 0 1 2-2

-1

0

1

2

x1

x2

m

θ=π/3

Example of Hebb in 2D

w

(Note: here inputs have a mean of zero)

The + symbols represent randomly chosen input pairs (x1,x2)

and the red circles the evolution of the synaptic weights

(w1=w2).

Part I: examples of ODE’s how to set them up and how to

solve them.

Example 1: Radioactive decay:

This is the canonical example for a simple 1D ODE, it is also a

good example for a random statistical processes, like you saw

yesterday

Assume there is a given amount or radioactive material defined

by the variable X. X(t) is the amount at time t.

The probability that each atom will decay in a small time period

∆t is independent of what the other molecules do, and only

radioactive molecules can decay to a non radioactive state. The

probability of decay over a small period ∆t is defined as γ·∆t.The

amount remaining at time t+∆t is:

Lets run the following simple matlab program to see what will

happen to x(t).

ttxtxttx ∆−=∆+ γ)()()(

>> x=zeros(1,1001);

>> dt=1/1000;

>> t=0:dt:1000*dt;

>>

>> x(1)=10;

>> gamma=2;

>>for jj=1:(length(t)-1)

x(jj+1)=x(jj)-dt*gamma*x(jj);

end

>> plot(t,x)

What is the shape of these curves?

How do they depend on the parameter γ?

This is a difference equation:

A little simple math:

Now assume that the time step ∆t approached 0 (is very small)

This is now a differential equation:

ttxtxttx ∆−=∆+ γ)()()(

γ

γ

)()()(

)()()(

txt

txttx

ttxtxttx

−=∆−∆+

∆−=−∆+

γxdt

dx

t

txttx

t−==

∆−∆+

→∆

)()(lim

0

xdt

dxγ−=

xdt

dxγ−= How do we solve this ODE?

Make a guess, assume that:

)exp()( tBAtx ⋅⋅=

Note for this choice of x(t):

Insert this back into the ODE above, get:

Which is a solution if B=-γ.

So:

)()exp( txBBtABdt

dx⋅=⋅⋅=

)()( txtxB ⋅−=⋅ γ

)/1exp()0()exp()( τγ −=⋅−⋅= xtAtx

Solution:

Where τ the time constant is

τ =1/γ, and the initial condition determines the other

Free constant x(0)=A

Terminology, this is a first order (only first derivatives) linear

differential equation. The dynamical variable x depends on

only one parameter, t. If it depends on additional paramerters

we might obtain partial differential equations, which we will

not discuss here.

)/1exp()0()( τ−= xtx

Example 2: Chemical reactions. Assume that when a

molecule

Of type A binds to a molecule of type B they can form a

product of type C. Denote as A+B C.

A represents the concentration of type A etc.

Assume now that the probability that type A will bind with

be depends on their concentration. Then:

Where γ is a rate constant.

BAdt

dC

BAdt

dA

BAdt

dA

⋅+=

⋅−=

⋅−=

γ

γ

γ

)3(

)2(

)1(

These are three coupled ordinary

differential equations}

BAdt

dC

BAdt

dA

BAdt

dA

⋅+=

⋅−=

⋅−=

γ

γ

γ

)3(

)2(

)1(

Note that there are

conservation equations:

A+C=Atot and B+C=Btot

len=1000

dt=1/len;

A=zeros(1,1000);

B=zeros(1,1000);

C=zeros(1,1000);

gamma=2;

A(1)=2;

B(1)=10;

timeline=0:dt:dt*len;

for ii=1:len

A(ii+1)=A(ii)-dt*gamma*A(ii)*B(ii);

B(ii+1)=B(ii)-dt*gamma*A(ii)*B(ii);

C(ii+1)=C(ii)+dt*gamma*A(ii)*B(ii);

end

plot(timeline,A,'b'); hold on;

plot(timeline,B,'r-.');plot(timeline,C,'k');

Matlab program

BAdt

dC

BAdt

dA

BAdt

dA

⋅+=

⋅−=

⋅−=

γ

γ

γ

)3(

)2(

)1(

BAdt

dC

BAdt

dA

BAdt

dA

⋅+=

⋅−=

⋅−=

γ

γ

γ

)3(

)2(

)1( Simplification.

Lets assume a case where B>>A.

The smallest value on B possible is

B(0)-A(0) which is close to B(0).

Replace then the dynamical variable B

with the parameter B=B0=B(0). Use

conservation A+C=A(0), get:

AAC

ABdt

dA

−=

−=

)0()2(

)1( 0γ

))exp(1)(0(

)exp()0(

0

0

tBAC

tBAA

γ

γ

−−=

−=

Obtain solution:

))exp(1)(0()(

)exp()0()(

0

0

tBAtC

tBAtA

γ

γ

−−=

−= + signs represent

the approximate

analytical solutions

A(0)=1, B(0)=10, γ=2

Play with the program

parameters and initial

conditions and

compare to the

analytical solution, see

when the approximate

solutions are no longer

a good approximation.

Fixed points when:

Is this fixed point stable, that is when we slightly move away

from the FP will the dynamics return us to the FP or take us

away from it?

{CkkC

k

B

k

ABdt

dC21

2

0

1

00 −=−= γγ321

Fixed points and their stability:

The problems until now are very simple and exactly, most

problems are not. Lets take a simple problem and pretend it is

not to see what we would do in such a case. Same problem

rewritten:

0

1

20 Ak

kC

dt

dC==⇒=

dt

dC

C

Here, the flow is toward

the FP, and the FP is

stable. If the line had a

positive slope, at the FP

it would be unstable.

1k

12 / kk

Terms used:

1. First order, only first derivatives

The equation is second order. It too has

a simple solution try inserting .

2. This equation is linear because it does not have terms of the

form .

Non linear equation are usually hard to solve exactly, and we

usually resort to finding their fixed points and the stability of these

fixed points.

dt

dx

02

2

=++ bdt

dxa

dt

xd

)exp()( tAtx λ⋅=

n

dt

dx

3. An equation of the form

Is non-homogeneous we have methods for solving such equations.

Textbooks for differential equations:

1. Elementary differential equations and boundary value problems.

Boyce and DiPrima. This has mostly analytical solutions and

background on fixed points and their stability.

To see examples of equations for specific problems solved in terms

of the fixed points and their stability you can look at:

2. Nonlinear Dynamics and Chaos: With Applications to Physics,

Biology, Chemistry and Engineering. S. Stogatz.

This is not a differntial equations textbook, but the methods used and

examples are very useful and it is an easy read.

0)(2

2

=++ xbdt

dxa

dt

xd

Recap – what did we learn until now?

Part II: The Hebb rule, an example of learning dynamics and

how we can solve a 2D example.

The Hebb rule:

Where:

If we insert into equation above we get:

yxtw ii η⋅∆=∆ x1 x2 x3 x4

w1 w2 w3 w4

y

yxdt

dw

t

wi

ii η=→∆∆

i

i

ixwy ∑=

j

j

iji xxw

dt

dw ∑=η

Simple 1D example (x1=x).

a. The input is constant over time: x=a

b. The input has a probability of ½ of being 1 and a probability of

½ of being -1. Assume learning is slow so that can take average

over input distribution:

x1

w1

y

wawxdt

dw22 ηη ==

)exp()0()( 2tawtw η=

( ) wwxwdt

dwηηη =−⋅+⋅==

444 3444 211

)1(5.0)1(5.0 222

)exp()0()( twtw η=

x1

w1

y)exp()0()( 2tawtw η=

( ) wwxwdt

dwηηη =−⋅+⋅==

444 3444 211

)1(5.0)1(5.0 222

)exp()0()( twtw η=

2D example.

The Hebb rule:

Average:

Matrix notation:

Assume:

x1 x2

w1 w2

y

j

j

iji xxw

dt

dw ∑=η

ij

j

j

j

jiji Qwxxw

dt

dw ∑∑ == ηη

wQw

η=dt

d

=

=

=

5.00.1

5.0

5.05.0

0.1

2

1

pwith

pwith

x

x

Show simple

matlab

simulation of

this example

where:

Assume

=

=

=

5.00.1

5.0

5.05.0

0.1

2

1

pwith

pwith

x

x

ij

j

ji Qw

dt

dw ∑= jiij xxQ =

=

⋅

⋅+

⋅

⋅=

625.05.0

5.0625.0

15.01

15.05.05.0

5.015.0

5.0115.0

2

2

2

2

Q

wQw

η=dt

d

x1

x2

wQw

η=dt

dLets pretend now

that we can simply

drop this average

symbol

Qww

η=dt

d

iii uuQ λ=⋅Find the eigen vectors of Q:

>> Q=[0.625, 0.5

0.5, 0.625];

>> [U,lam]=eig(Q)

U =

-0.7071 0.7071

0.7071 0.7071

lam =

0.1250 0

0 1.1250

Matlab code

125.01

1

2

1

125.11

1

2

1

22

11

=

−

=

=

=

λ

λ

u

u

Qww

η=dt

d

iii uuQ λ=⋅Find the eigen-vectors of Q:

These form a complete orthonormal basis

Rewrite:

So

21 uuw(t) )()( 21 tata +=

( )22211 )()( uuw

1 λλη tatadt

d+=

iii a

dt

daηλ=

21 uuw(t) )exp()0()exp()0( 2211 tata λλ +=

The averaged

‘Hebbian’ ODE:

Qww

η=dt

d

iii uuQ λ=⋅Find the eigen-vectors of Q:

If Q has the common form:

Then:

What happens if λ1>> λ2 ?

21 uuw(t) )exp()0()exp()0( 2211 tata λλ +=

The averaged

‘Hebbian’ ODE:

=

ab

baQ

−

+=−=

+

+=+=

1

1

2

1

1

1

2

1

11

11

uba

uba

λ

λ

General comments about the form of correlation matrixes.

• They are symmetric Qij=Qji., and therefore all of their eigen-

values are real.

• All of their eigen-values are positive (Try to prove this yourself)

HW- For the general form of a correlation matrix in 2D:

1. Learn how to find analytically the eigen-vectors and

eigen-values.

2. Show that in this case all eigen-values are positive

3. Show that in the higer dimnstional case all eigen-values

are positive

=

bc

caQ

What did we learn today?

HW-

What would happen with the learning rule:

Oja (1982)

Where are the F.P, how does this relate to the eigen-

vectors, and why

( )2ywyxdt

dwii

i −=η